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Motion in a Straight Line & Plane

Master Motion in a Straight Line & Plane for JEE Main, JEE Advanced, NEET & Board Exams. Covers kinematics equations, projectile motion, relative motion, free fall, river-boat problems & more with solved examples.

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Motion is the change in position of an object with respect to time. When we study motion along a fixed direction, it is called motion in a straight line (rectilinear motion). When motion occurs in two dimensions simultaneously — such as a ball thrown at an angle — it is called motion in a plane. Both are governed by the equations of kinematics, which relate displacement, velocity, acceleration, and time. This topic is foundational for JEE (Main & Advanced), NEET, and all board exams, and forms the backbone of classical mechanics.

1. Basic Definitions and Key Quantities

(i) Distance vs. Displacement

Distance is the total path length travelled by an object. It is a scalar quantity and is always non-negative.
Displacement ( $\vec{s}$ ) is the shortest straight-line distance from the initial to the final position, along with direction. It is a vector quantity and can be zero, positive, or negative.

For any motion: $| Displacement | \leq Distance$ . Equality holds only when the object moves in a single straight line without reversing direction.

(ii) Speed vs. Velocity

Quantity	Formula	Nature	Can be negative?
Average Speed	$\frac{Total Distance}{Total Time}$	Scalar	No
Average Velocity	$\frac{Δ \vec{x}}{Δ t} = \frac{{\vec{x}}_{f} - {\vec{x}}_{i}}{t}$	Vector	Yes
Instantaneous Speed	$\| \frac{d x}{d t} \|$	Scalar	No
Instantaneous Velocity	$\vec{v} = \frac{d \vec{x}}{d t}$	Vector	Yes

(iii) Acceleration

Acceleration is the rate of change of velocity with time:

$\vec{a} = \frac{d \vec{v}}{d t} = \frac{d^{2} \vec{x}}{d t^{2}}$

Acceleration is a vector. It can be positive, negative (retardation), or zero.
A body can have zero velocity but non-zero acceleration (e.g., at the highest point of a vertical throw).
A body can have constant speed but non-zero acceleration (e.g., uniform circular motion).

2. Equations of Motion for Uniform Acceleration

When a body moves along a straight line with constant acceleration $a$ , the following three kinematic equations apply. Here, $u$ = initial velocity, $v$ = final velocity, $s$ = displacement, $t$ = time.

Equation	Quantities NOT involved
$v = u + a t$	$s$
$s = u t + \frac{1}{2} a t^{2}$	$v$
$v^{2} = u^{2} + 2 a s$	$t$
$s = \frac{(u + v)}{2} \cdot t$	$a$

Displacement in the $n^{th}$ Second

The displacement of a uniformly accelerating body specifically in the $n^{th}$ second (not in $n$ seconds) is given by:

$s_{n} = u + \frac{a}{2} (2 n - 1)$

This formula is very frequently tested in JEE and NEET. Note that $s_{n}$ is a displacement (not distance) and can be negative.

Graphical Interpretation of Motion

Displacement–Time ( $x$ – $t$ ) graph: The slope at any point gives the instantaneous velocity. A straight line means constant velocity; a curve means changing velocity (acceleration). A horizontal line means the body is at rest.
Velocity–Time ( $v$ – $t$ ) graph: The slope at any point gives the instantaneous acceleration. The area under the $v$ – $t$ graph between two time points gives the displacement in that interval.
Acceleration–Time ( $a$ – $t$ ) graph: The area under the $a$ – $t$ graph gives the change in velocity.

Important: For a $v$ – $t$ graph, the area above the time-axis is positive displacement and the area below is negative displacement. Total distance = sum of magnitudes of both areas.

3. Motion Under Gravity (Free Fall)

Near the Earth's surface, all freely falling bodies experience the same downward acceleration due to gravity, $g \approx 9.8 {m/s}^{2} \approx 10 {m/s}^{2}$ (for calculations). Taking upward as positive:

Situation	Initial velocity $u$	Acceleration $a$
Object dropped from rest	$0$	$- g$
Object thrown upward	$+ u$	$- g$
Object thrown downward	$- u$	$- g$

Key Results for a Body Thrown Vertically Upward with speed $u$ :

Time to reach maximum height: $t_{u p} = \frac{u}{g}$
Maximum height reached: $H = \frac{u^{2}}{2 g}$
Total time of flight (back to same level): $T = \frac{2 u}{g}$
Speed on return to the same level = $u$ (same as initial speed, opposite direction)
The motion is symmetric about the highest point — time going up equals time coming down.

4. Relative Motion in a Straight Line

The velocity of object A as observed from object B is called the velocity of A relative to B:

${\vec{v}}_{A B} = {\vec{v}}_{A} - {\vec{v}}_{B}$

If two objects move in the same direction with speeds $v_{A}$ and $v_{B}$ : relative speed $= | v_{A} - v_{B} |$
If two objects move in opposite directions: relative speed $= v_{A} + v_{B}$

Similarly, relative acceleration: ${\vec{a}}_{A B} = {\vec{a}}_{A} - {\vec{a}}_{B}$

For two objects under gravity (same $g$ downward), $a_{A B} = g - g = 0$ , so they appear to move with constant velocity relative to each other. This is why astronauts in a spacecraft appear weightless.

5. Motion in a Plane — Vectors in 2D

Motion in a plane is described using two-dimensional vectors. Position, velocity, and acceleration are all vector quantities resolved into perpendicular components (usually $x$ and $y$ ).

$\vec{r} = x \hat{i} + y \hat{j}$ , $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j}$ , $\vec{a} = a_{x} \hat{i} + a_{y} \hat{j}$

The key principle is that horizontal and vertical motions are completely independent. Each component is analysed separately using the equations of motion.

Component	Velocity	Position
$x$ -direction (horizontal)	$v_{x} = v_{x 0} + a_{x} t$	$x = x_{0} + v_{x 0} t + \frac{1}{2} a_{x} t^{2}$
$y$ -direction (vertical)	$v_{y} = v_{y 0} + a_{y} t$	$y = y_{0} + v_{y 0} t + \frac{1}{2} a_{y} t^{2}$

6. Projectile Motion

A projectile is any object launched into space with only gravity acting on it (air resistance neglected). The path traced is a parabola. If launched with speed $u$ at angle $θ$ above the horizontal:

Component	Initial value	Acceleration	At time $t$
Horizontal ( $x$ )	$u_{x} = u \cos θ$	$0$ (constant)	$x = u \cos θ \cdot t$
Vertical ( $y$ )	$u_{y} = u \sin θ$	$- g$ (downward)	$y = u \sin θ \cdot t - \frac{1}{2} g t^{2}$

Important Formulae

Time of Flight: $T = \frac{2 u \sin θ}{g}$
Maximum Height: $H = \frac{u^{2} \sin^{2} θ}{2 g}$
Horizontal Range: $R = \frac{u^{2} \sin 2 θ}{g}$
Maximum Range: $R_{m a x} = \frac{u^{2}}{g}$ when $θ = 45 °$
Equation of Trajectory (path): $y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$

At any point during flight, the speed is $v = \sqrt{v_{x}^{2} + v_{y}^{2}}$ and the angle with horizontal is $ϕ = \tan^{- 1} (\frac{v_{y}}{v_{x}})$ . The horizontal component $v_{x} = u \cos θ$ remains constant throughout.

Complementary Angles and Projectile Range

Two projection angles that are complementary (i.e., $θ$ and $90 ° - θ$ ) give the same horizontal range but different maximum heights and times of flight.

$R (θ) = R (90 ° - θ)$ , since $\sin 2 θ = \sin (180 ° - 2 θ)$ .
The ratio of maximum heights: $\frac{H_{1}}{H_{2}} = \frac{\sin^{2} θ}{\cos^{2} θ} = \tan^{2} θ$
The ratio of times of flight: $\frac{T_{1}}{T_{2}} = \frac{\sin θ}{\cos θ} = \tan θ$
Useful result: $R = \frac{4 H}{\tan θ}$ , linking range and max height.

7. Relative Motion in a Plane — River-Boat & Rain Problems

(i) River-Boat Problem

A boat of speed $v_{b}$ (in still water) crosses a river of width $d$ flowing with speed $v_{r}$ . The velocity of the boat with respect to the ground is:

${\vec{v}}_{b o a t, g r o u n d} = {\vec{v}}_{b o a t, r i v e r} + {\vec{v}}_{r i v e r, g r o u n d}$

To minimize time of crossing: Aim the boat perpendicular to the river. Time $= \frac{d}{v_{b}}$ . Drift $= v_{r} \cdot \frac{d}{v_{b}}$ .
To minimize drift (shortest path): Aim the boat at angle $\sin^{- 1} (\frac{v_{r}}{v_{b}})$ upstream. This is possible only if $v_{b} > v_{r}$ . Drift = 0, and time $= \frac{d}{\sqrt{v_{b}^{2} - v_{r}^{2}}}$ .

(ii) Rain Problem

A person walking with velocity ${\vec{v}}_{m}$ in rain falling with velocity ${\vec{v}}_{r}$ must hold the umbrella in the direction of the relative velocity of rain with respect to the person:

${\vec{v}}_{r, m} = {\vec{v}}_{r} - {\vec{v}}_{m}$

The angle of the umbrella with vertical is $θ = \tan^{- 1} (\frac{v_{m}}{v_{r}})$ (tilted forward in the direction of motion).

8. Non-Uniform Acceleration — Calculus Approach

When acceleration is not constant, the three kinematic equations do not apply directly. Instead, use calculus:

$v = \frac{d x}{d t}$ ⟹ $d x = v d t$ ⟹ $x = \int v d t$
$a = \frac{d v}{d t}$ ⟹ $d v = a d t$ ⟹ $v = \int a d t$
$a = v \frac{d v}{d x}$ ⟹ $v d v = a d x$ ⟹ useful when $a$ is given as a function of $x$ .

Example: If $a = 3 t^{2} - 2 t$ and $v = 0$ at $t = 0$ , then $v = \int_{0}^{t} (3 t^{2} - 2 t) d t = t^{3} - t^{2}$ .

JEE/NEET Power Tips — Avoid These Common Mistakes

Speed ≠ Magnitude of average velocity. Average speed uses total distance; average velocity uses displacement.
$s_{n}$ formula gives displacement, not distance. If the object reverses direction within the $n^{th}$ second, this formula can give a misleading result for distance.
At the highest point of projectile motion, only the vertical velocity $v_{y} = 0$ , NOT the total velocity. The horizontal velocity $v_{x} = u \cos θ$ is still present.
Range is maximum at $θ = 45 °$ only when projected from and landing on the same horizontal level. For inclined planes or different heights, the optimum angle changes.
In relative motion problems, always set up vectors with a consistent sign convention before subtracting. Drawing a vector diagram first saves errors.
Acceleration due to gravity is constant only near the Earth's surface. Do not apply $g = 9.8$ m/s² at very large heights.

Practice Questions (JEE / NEET / Board Level)

Q1: A ball is thrown vertically upward with a speed of 20 m/s. Taking $g = 10 {m/s}^{2}$ , which of the following is correct at $t = 3 s$ ?

A) The ball is at maximum height.
B) The ball has returned to the ground.
C) The ball is moving downward with speed 10 m/s.
D) The ball is at height 25 m above the launch point.

Answer: C) The ball is moving downward with speed 10 m/s.

Explanation: First, find the time it takes to reach maximum height using $v = u - g t$ . Setting $v = 0$ , we get $t = \frac{20}{10} = 2 s$ . Since we are looking at $t = 3 s$ , the ball has passed its peak and is descending.

To find the exact velocity at $t = 3 s$ , use the first equation of motion:
$v = u - g t = 20 - 10 (3) = - 10 m/s$ .
The negative sign indicates the ball is moving downward with a speed of 10 m/s.

Note: To verify its height at $t = 3 s$ , use $s = u t - \frac{1}{2} g t^{2}$ .
$s = 20 (3) - \frac{1}{2} (10) (3^{2}) = 60 - 45 = 15 m$ . Therefore, option D is incorrect.

Q2: A particle's displacement is given by $x = 3 t^{3} - 4 t^{2} + 2 t + 5$ (in metres, $t$ in seconds). What is its acceleration at $t = 2$ s?

A) $18$ m/s²
B) $28$ m/s²
C) $22$ m/s²
D) $10$ m/s²

Answer: B) 28 m/s². $v = \frac{d x}{d t} = 9 t^{2} - 8 t + 2$ . $a = \frac{d v}{d t} = 18 t - 8$ . At $t = 2$ : $a = 18 (2) - 8 = 36 - 8 = 28$ m/s².

Q3: A projectile is fired with speed $u = 20$ m/s at $θ = 30 °$ above the horizontal ( $g = 10$ m/s²). What is the horizontal range?

A) $20 \sqrt{3}$ m
B) $40$ m
C) $20$ m
D) $10 \sqrt{3}$ m

Answer: A) $20 \sqrt{3}$ m.
$R = \frac{u^{2} \sin 2 θ}{g}$
$= \frac{(20)^{2} \sin 60 °}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10}$
$= \frac{200 \sqrt{3}}{10} = 20 \sqrt{3}$ m.

Q4: A boat can travel at $5$ m/s in still water. A river flows at $3$ m/s. The minimum time to cross a $120$ m wide river is:

A) $24$ s
B) $30$ s
C) $40$ s
D) $20$ s

Answer: A) $24$ s. Minimum time requires the boat to be aimed perpendicular to the river banks. Time $= \frac{d}{v_{b}} = \frac{120}{5} = 24$ s. (The river's current does not affect crossing time, only the drift.)

Q5: The velocity–time graph of a particle moving along a straight line is a straight line with a positive slope passing through the origin. This implies:

A) Constant velocity
B) Uniformly increasing velocity, starting from rest
C) Uniformly decreasing velocity
D) Non-uniform acceleration

Answer: B) A straight $v$ – $t$ graph through the origin means $v \propto t$ , i.e., $v = a t$ where $a$ is constant. The body starts from rest ( $v = 0$ at $t = 0$ ) and accelerates uniformly.

Q6 (JEE Advanced type): Two particles A and B are projected simultaneously from the same point. A is projected horizontally with speed $10$ m/s and B is projected vertically upward with speed $20$ m/s. After $2$ s, what is the magnitude of the relative velocity of B with respect to A?

A) $10 \sqrt{5}$ m/s
B) $10 \sqrt{2}$ m/s
C) $20$ m/s
D) $10$ m/s

Answer: A) $10 \sqrt{5}$ m/s. At $t = 2$ s: $v_{A} = 10 \hat{i} - 20 \hat{j}$ m/s (horizontal + gravity); $v_{B} = 0 \hat{i} + (20 - 20) \hat{j} = 0 \hat{j}$ m/s (vertical, at max height). ${\vec{v}}_{B A} = {\vec{v}}_{B} - {\vec{v}}_{A} = - 10 \hat{i} + 20 \hat{j}$ m/s. $| {\vec{v}}_{B A} | = \sqrt{100 + 400} = \sqrt{500} = 10 \sqrt{5}$ m/s.

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Motion in a Straight Line & Plane

1. Basic Definitions and Key Quantities

(i) Distance vs. Displacement

(ii) Speed vs. Velocity

(iii) Acceleration

2. Equations of Motion for Uniform Acceleration

Displacement in the $n^{th}$ Second

Graphical Interpretation of Motion

3. Motion Under Gravity (Free Fall)

Key Results for a Body Thrown Vertically Upward with speed $u$ :

4. Relative Motion in a Straight Line

5. Motion in a Plane — Vectors in 2D

6. Projectile Motion

Important Formulae

Complementary Angles and Projectile Range

7. Relative Motion in a Plane — River-Boat & Rain Problems

(i) River-Boat Problem

(ii) Rain Problem

8. Non-Uniform Acceleration — Calculus Approach

Practice Questions (JEE / NEET / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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Edvaya Target

Edvaya Aspire

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Target Subjects

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Subject Mastery

Motion in a Straight Line & Plane

1. Basic Definitions and Key Quantities

(i) Distance vs. Displacement

(ii) Speed vs. Velocity

(iii) Acceleration

2. Equations of Motion for Uniform Acceleration

Displacement in the nth Second

Graphical Interpretation of Motion

3. Motion Under Gravity (Free Fall)

Key Results for a Body Thrown Vertically Upward with speed u:

4. Relative Motion in a Straight Line

5. Motion in a Plane — Vectors in 2D

6. Projectile Motion

Important Formulae

Complementary Angles and Projectile Range

7. Relative Motion in a Plane — River-Boat & Rain Problems

(i) River-Boat Problem

(ii) Rain Problem

8. Non-Uniform Acceleration — Calculus Approach

Practice Questions (JEE / NEET / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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We Value Your Privacy

Displacement in the $n^{th}$ Second

Key Results for a Body Thrown Vertically Upward with speed $u$ :